# Idiom #81 Round floating point number to integer

Declare the integer y and initialize it with the rounded value of the floating point number x .
Ties (when the fractional part of x is exactly .5) must be rounded up (to positive infinity).

y = int(x + 0.5)
Y : constant Integer := Integer (Float'Rounding (X));
#include <math.h>
int y = (int)floorf(x + 0.5f);
(defn rnd [y] (int (+ y 0.5)))
#include <cmath>
int y = static_cast<int>(std::floor(x + 0.5f));
using System;
long y = (long)Math.Round(x);
import std.math: round;
int y = cast(int) x.round;
var y = x.round();
y = Kernel.round x
integer :: y

y = nint(x)
import "math"
y := int(math.Floor(x + 0.5))
y = floor (x + 1/2)
var y = Math.round(x);
long y = Math.round(x);
(defun rnd (y) (floor (+ y 0.5)))
function round(float)
local int, part = math.modf(float)
if float == math.abs(float) and part >= .5 then return int+1    -- positive float
elseif part <= -.5 then return int-1                            -- negative float
end
return int
end
function round(float)
return math.floor(float + .5)
end
\$y = (int) round(\$x);
var
y: integer;
x: double;
begin
y := round(x);
end.
my \$y = int(\$x + 1/2);
y = (x + 1/2r).floor
let y = x.round() as i64;
(define y (round x))

deleplace